In the book of Kondepudi, Modern Thermodynamics, at page 28, it is given that

According to Boltzmann principle, the probability that a molecule will have a translational energy $E_{trans}$ is proportional to $e^{-E_{trans}/(k_B t)}$. Since the kinetic energy due to translational motion of the molecule is $mv2/2$, we can write the probability as a function of the velocity v by which we mean probability that a molecule’s velocity is in an elemental cube in velocity space. For a continuous variable, such as velocity, we must define a probability density $P(v)$ so that the probability that a molecule’s velocity is in an elemental cube of volume $dv_x dv_y dv_z$ located at the tip of the velocity vector $v$ is $P(v) dv_x dv_y dv_z$. According to the Boltzmann principle, this probability is

$$P(\mathbf{v}) \mathrm{d} v_{x} \mathrm{d} v_{y} \mathrm{d} v_{z}=\frac{1}{z} \mathrm{e}^{-m v^{2} / 2 k_{\mathrm{B}} T} \mathrm{d} v_{x} \mathrm{d} v_{y} \mathrm{d} v_{z}$$

However, I'm having hard time understanding why $P(v)$ is given as above.

According to Boltzmann principle, (lets forget the subscript "trans" for the time being) $$P(E) dE \propto \exp{(-E/(k_BT))},$$ but since $E \propto mv^2 /2,$

$$P(mv^2 /2)mvdv \propto \exp{ (-mv^2 / (2k_B T))}.$$

but this can be written as

$$P(v^2) v dv \propto \exp{ (-mv^2 / (2k_B T))}.$$

However, at this point, even if I make some symmetry arguments, I'm not able to reach the same conclusion as the author does.


How can we derive the given velocity distribution in the book from the argument that the author provides ?

  • $\begingroup$ What does the $z$ represent in the cited text? $\endgroup$
    – WarreG
    Commented Aug 18, 2019 at 21:10
  • $\begingroup$ @WarreG $z$-direction, like $x$ and $y$. $\endgroup$
    – Our
    Commented Aug 19, 2019 at 3:19

2 Answers 2


I recommend to abandon this argument based on $P(E)dE$, which I assume means probability that energy is between $E,E+dE$. Boltzmann's distribution law is not primarily about probability of system having some energy $E$ (but of course, this can be derived from it).

What Boltzmann's law says is that given discrete possible states $i$ of the system, probability that the system is in state $i$ depends only on energy of that state $E_i$ and is equal to

$$ P_i = \frac{e^{-\frac{E_i}{k_BT}}}{Z}. $$

To put it simply, Boltzmann's law is about probability of states, not probability of energy.

When the states form a continuum, such as state defined by position of particle in space and its velocity, the above formula can't be used; there is too many states and probability of single state has to be zero.

The formulation has to be modified; usually we use distribution function in space and velocity space, so that the above formula gives probability density that has to be integrated over region of space and velocity space to obtain probability.

Let energy of molecule with position $\mathbf r$ and velocity $\mathbf v$ be

$$ E = \frac{1}{2}mv^2+U(\mathbf r) $$ The Boltzmann's probability distribution for molecule position and velocity is

$$ \rho(\mathbf r, \mathbf v) = \frac{e^{-\frac{\frac{1}{2}mv^2+U(\mathbf r)}{k_BT}}}{Z}. $$

From this, using substitution theorem on integrals you can find probability density for energy, the function $P(E)$ you mentioned in your question.


Your textbook is a bit imprecise. The proper way to state the Boltzmann probability is:

the probability to find a particle with velocity between $\mathbf{v}$ and $\mathbf v+d\mathbf{v}$ is proportional to $e^{-\frac{m}{2kT}(v_x^2+v_y^2+v_z^2)} dv_x dv_y dv_z$.

The problem here is that $e^{-E/kT}$ is not the probability to find a particle with energy $E$; it is the probability to find it at the microstate $\mathbf v = (v_x,v_y,v_z)$.

  • $\begingroup$ but since $E\propto v^2$, up a factor counting the # of microstate for that giden macrostate, the same result stili applies. $\endgroup$
    – Our
    Commented Aug 19, 2019 at 3:16

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